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If the radius of earth is R then height ‘h’ at which value of ‘g’ becomes one - fourth is
$ g' = g \left( { R \over R+ h } \right)^2 = 9/4 by solving \, h = R $
If the mass of earth is 80 times of that of a planet and diameter is double that of planet and ‘g’ on the earth is $9.8 ms^{-2}$ , then the value of ‘g’ on that planet is = ............... $ms^{-2}$
$g_p = g_e \left( { M_p \over M_e } \right) = \left( { R_e \over R_p } \right)^2 = 9.8 \left( { 1 \over 80} \right)(2)^2 = { 9.8 \over 20 } = 0.49 ms^{-2} $
Assuming earth to be a sphere of a uniform density, what is value of gravitational acceleration in mine 100 km below the earth surface = ............... $ms^{-2}$
$ g' = g \left( { 1 - { d \over R }} \right) = 9.8 \left( { 1 - {100 \over 6400} } \right) = 9.66 ms^{-2} $
Let g be the acceleration due to gravity at earth’s surface and k be the rotational K.E. of earth suppose the earth’s radius decreases by 2% keeping alt other quantities same then
$ g = { GM \over R^2 } $ and $ k = { L^2 \over 2I } $ If mass of the earth and its angular momentum ramains constant then$ g \alpha { 1 \over R^2 } and \;k \alpha { 1 \over R^2 } $ i.e. if radius of earth decreases by 2% then g and k both increase by 4%
At what height over the earth’s pole, the free fall acceleration decreases by one percent = .................. km (Re = 6400 km).
$ g \alpha { GM \over r^2 } $ $ \therefore g \alpha { 1 \over r^2 } $ or $r \alpha { 1 \over \sqrt g }$ If g decreases by one percent then r should be increase by 1/2 % i.e. $ R = {1 \over 2 \times 100 } \times 6400 = 32 km $
Weight of a body is maximum at
At what distance from the center of earth, the value of aceeleration due to gravity g will be half that of the surfaces (R = Radius of earth)
$ g' = g \left( R \over R+h \right) ^2 \Rightarrow { 1 \over \sqrt 2 } = { R \over R+h } \Rightarrow R+h \Rightarrow \sqrt 2 R \Rightarrow h = (\sqrt { 2-1} )R = 0.414 R $ Hence distance form center = R + 0.414R = 1.414R
The acceleration due to gravity near the surface of a planet of radius R and density d is proportional to
$ g = { 4 \over 3 } \pi \rho GR \Rightarrow g \alpha d R ( \rho = d given in the problem ) $
The acceleration due to gravity is g at a point distance r from the center of earth R. if r < R then
inside the earth $ g ' = {4 \over 3 } \pi Gr \therefore g' \alpha r $
Density of the earth is doubled keeping its radius constant then acceleration, due to gravity will be .................. $ ms^{-2}(g = 9.8 ms^2)$
$ g \alpha \rho $
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The bank mixes NEET previous year questions (PYQs) with practice questions, each tagged with its exam appearances where applicable.